How Far Is a Sample Mean from the Center?
In “Reading a Simulated Sampling Distribution of x-bar,” you interpreted how sample means vary across repeated samples. Now we can measure the position of one observed sample mean relative to that distribution. The key is to compare its distance from the population mean with the typical distance of sample means from that mean.
As established in “Mean of the Sampling Distribution of x-bar” and “Standard Deviation of the Sample Mean,” the sampling distribution of \(\bar{x}\) is centered at \(\mu\), and, for independent observations, its standard deviation is \(\sigma/\sqrt{n}\). This standard deviation tells us the typical distance between sample means and \(\mu\). Dividing a sample mean’s distance from \(\mu\) by that amount puts the distance on a common, unitless scale.
A positive z-score means \(\bar{x}\) is above \(\mu\); a negative z-score means it is below \(\mu\). For example, \(z=1.5\) means the sample mean is 1.5 standard deviations of the sampling distribution above \(\mu\). It does not mean 1.5% or 15%.
The denominator is the standard deviation of sample means, not the standard deviation of individual observations. It is often called the standard error of \(\bar{x}\), though “standard deviation of the sampling distribution” is precise when \(\sigma\) is known. The numerator and denominator have the same units, so the z-score has no units.
When a Z-Score Helps Judge Unusualness
A z-score describes distance from the center, but judging unusualness also requires knowing the shape of the sampling distribution. If the population is Normal, the sampling distribution of \(\bar{x}\) is Normal for any sample size, provided the observations are independent. If the population is not Normal, the Central Limit Theorem can make the sampling distribution approximately Normal for a sufficiently large random sample. As discussed in “Is n = 30 Large Enough for the CLT,” the adequacy of a sample size depends on the population’s shape; 30 is not a guarantee for every population.
When the sampling distribution is Normal or approximately Normal, convert the observed \(\bar{x}\) to a z-score and find the corresponding tail area. A very small tail area indicates that a sample mean at least that far in the specified direction would be uncommon if the population mean and standard deviation were as stated. For a two-sided question about unusually low or high means, consider sample means at least as far from \(\mu\) in either direction.
A common rule of thumb is that a value more than about two standard deviations from the center is unusual. For a standard Normal distribution, the probability of being at least two standard deviations above the mean is about 0.0228, and the probability of being at least two standard deviations away in either direction is about 0.0455. These are small probabilities, but “unusual” is a judgment rather than a universal cutoff. Look at the relevant tail area and the situation, not only whether a rounded z-score crosses 2.
Worked Examples: Calculating and Interpreting Z-Scores
Worked Example: A Sample Mean from a Normal Population
A machine fills containers. For this invented example, suppose individual fill amounts follow a Normal distribution with mean \(\mu=500\) milliliters and standard deviation \(\sigma=12\) milliliters. A random sample of \(n=36\) containers has mean fill amount \(\bar{x}=504\) milliliters. Is this sample mean unusual if the machine is operating according to the stated model?
State. We are assessing whether a sample mean of 504 mL is unusually high relative to the sampling distribution when \(\mu=500\) mL and \(\sigma=12\) mL.
Plan. The sample is random. The observations are independent under the stated sampling setup; if the containers are sampled without replacement from a finite production run, the 10% condition must also hold. The population is stated to be Normal, so the sampling distribution of \(\bar{x}\) is Normal. We can calculate its standard deviation and standardize the observed mean.
Do. The standard deviation of the sampling distribution is
Thus, 504 mL is 4 mL above the center, or \(4/2=2\) standard deviations above it:
As a check, the standard Normal score is also \(4\div2=2\). The upper-tail probability is \(P(Z\geq2)\approx0.0228\), rounded to four decimal places. If the question were whether a sample mean is unusual in either direction, the two-sided probability would be \(2P(Z\geq2)=2(0.022750\ldots)\approx0.0455\), using the unrounded tail probability.
Conclude. The observed mean is two standard deviations above the expected mean. Only about 2.28% of sample means would be at least this high under the stated Normal model, so 504 mL is unusual as a high sample mean. This does not prove that the machine is malfunctioning; it says the result is uncommon under the model.
Worked Example: A Large Sample from a Right-Skewed Population
Suppose the time customers spend waiting for a service is right-skewed, with population mean \(\mu=32\) minutes and standard deviation \(\sigma=20\) minutes. A random sample of \(n=100\) customers has mean waiting time \(\bar{x}=36.2\) minutes. Assess whether this sample mean is unusually high under the stated model.
State. We want to determine whether 36.2 minutes is unusually high for the mean wait of a sample of 100 customers.
Plan. Assume the 100 customers form a random sample and their wait times are independent. If sampled without replacement from a finite group of \(N\) customers, verify \(100<0.10N\). The population is right-skewed, not Normal, so we need the sample size and shape to support an approximately Normal sampling distribution. Here \(n=100\), and the population is stated to have a finite standard deviation; the Central Limit Theorem supports an approximately Normal sampling distribution only if the right skew is not so extreme that \(n=100\) is inadequate, and the information given is not enough to assess that adequacy. The approximation may be less accurate if the population is extremely skewed or has influential outliers.
Do. The standard deviation of the sample mean is
The observed sample mean is \(36.2-32=4.2\) minutes above \(\mu\). Dividing that difference by the standard deviation of sample means gives
The same result comes from expressing the difference in standard-error units: \(4.2\div2=2.1\). Using the Normal approximation, the upper-tail area is \(P(Z\geq2.1)\approx0.0179\), rounded to four decimal places. Equivalently, a calculator’s normalcdf with lower bound 36.2, a suitably large upper bound, mean 32, and standard deviation 2 gives approximately 0.0179.
Conclude. The sample mean is 2.1 standard deviations above the population mean. The approximate probability of a sample mean at least 36.2 minutes is 0.0179, or about 1.79%, under the stated model. This is unusual evidence of a high sample mean, with the conclusion qualified by the Normal approximation.
Worked Example: The Same Difference Can Have Different Z-Scores
A sensor’s readings are Normally distributed with mean \(\mu=70\) units and standard deviation \(\sigma=12\) units. Consider a sample mean of 74 units. Compare how unusual it would be for samples of \(n=9\) and \(n=36\).
State. In each case, the sample mean is 4 units above the population mean. We will compare its z-score and upper-tail probability for two sample sizes.
Plan. Assume independent random samples; if sampling without replacement from a finite population, check the 10% condition for each sample size. Since the population is Normal, each sampling distribution is Normal. Calculate \(\sigma/\sqrt{n}\) separately for each sample size, then standardize 74 using that sample mean’s own standard deviation.
Do. For \(n=9\), the standard deviation of sample means is \(12/\sqrt{9}=12/3=4\) units. Therefore,
The upper-tail probability is \(P(Z\geq1)\approx0.1587\), rounded to four decimal places. A mean of 74 is not especially unusual for samples of size 9.
For \(n=36\), the standard deviation of sample means is \(12/\sqrt{36}=12/6=2\) units. Now,
The upper-tail probability is \(P(Z\geq2)\approx0.0228\). The calculations can also be checked by dividing the same 4-unit difference by each standard deviation: \(4\div4=1\) for \(n=9\), and \(4\div2=2\) for \(n=36\).
Conclude. A sample mean of 74 is only one standard deviation above the center for \(n=9\), but two standard deviations above the center for \(n=36\). Its upper-tail probability falls from about 0.1587 to about 0.0228. The larger sample has less variability in its sample means, so the same 4-unit difference is more unusual. This comparison is about the sampling distributions; it does not claim that every larger sample will produce a mean closer to \(\mu\).
Common Mistakes and AP Exam Tip
- Using \(\sigma\) instead of \(\sigma/\sqrt{n}\): The formula concerns a sample mean, so its standard deviation is \(\sigma/\sqrt{n}\), not the spread of individual observations.
- Dropping the square root: Compute \(\sqrt{n}\) before dividing. For instance, if \(\sigma=12\) and \(n=36\), the standard deviation of \(\bar{x}\) is \(12/6=2\), not \(12/36\).
- Misreading the sign: A positive z-score means the observed sample mean is above \(\mu\); a negative z-score means it is below. Use the sign to match the question’s direction.
- Calling z a probability: A z-score is a distance measured in standard deviations, not a percent. To discuss how often the result would occur, find the appropriate tail probability under the Normal model.
- Using the wrong tail: For “unusually high,” find the area to the right. For “unusually low,” find the area to the left. For “unusually far from the mean,” account for both tails.
- Assuming every z-score can be judged with a Normal model: State why the sampling distribution is Normal or approximately Normal, and check randomization, independence, and the 10% condition when relevant.
- Claiming an unusual result proves a cause: An unusual sample mean is uncommon under the stated model; it does not, by itself, identify an explanation or prove the model is wrong.
For full-credit communication, show the standard deviation of the sampling distribution, substitute into the z-score formula, state the direction and number of standard deviations from \(\mu\), and interpret the relevant probability in context. Include the conditions that support the Normal model and describe an unusual result as uncommon under that model—not impossible.
Check Your Understanding
Use the formula and reasoning in this tutorial to answer the questions.
- A Normal population has \(\mu=40\), \(\sigma=10\), and a sample of \(n=25\) has \(\bar{x}=44\). Calculate the standard deviation of the sampling distribution and the z-score.
- In Question 1, is the sample mean unusually high using the approximate two-standard-deviation guideline? Explain what the z-score means in context.
- A right-skewed population has a finite standard deviation. What should you consider before treating the sampling distribution of \(\bar{x}\) as approximately Normal for a sample of \(n=12\)?
- For a Normal population with \(\mu=80\) and \(\sigma=15\), compare the standard deviations of sample means for \(n=9\) and \(n=36\). Which sample mean would make an observed value of 85 more unusual?
- Explain why an upper-tail probability of 0.01 does not mean that the observed sample mean has a 1% probability of being the true population mean.